Exercise · ZW-SEPARABLE-EQUATION-008 · ★★☆ Intermediate
When solving \(y'=y(y-2)\) we divide by \(y(y-2)\). Which constant solutions might be lost?
Choose one option.
Show answer
- D. \(y=0\) and \(y=2\)
Explanation
Dividing by \(h(y)=y(y-2)\) assumes \(h(y)\ne0\). The roots \(y=0\) and \(y=2\) make the right side zero, so they are constant solutions that must be added separately.
Review the concept: Separable equation
A separable equation is a first-order differential equation whose derivative is the product of a function \(g\) of \(x\) alone and a function \(h\) of \(y\) alone.
\[\frac{dy}{dx}=g(x)\,h(y)\;\Longrightarrow\;\int\frac{dy}{h(y)}=\int g(x)\,dx\]
More practice problems: separable equation
- ZW-SEPARABLE-EQUATION-001 Which of these equations is separable?
- ZW-SEPARABLE-EQUATION-002 When separating variables in y'=2xy, which equality of integrals results?
- ZW-SEPARABLE-EQUATION-003 Find the general solution of y'=3y.
- ZW-SEPARABLE-EQUATION-004 Separating variables in y'=x/y, which implicit relation do the solutions satisfy?
- ZW-SEPARABLE-EQUATION-005 Solve the problem y'=y, y(0)=2.
- ZW-SEPARABLE-EQUATION-006 Solve y'=-2xy, y(0)=3.
- ZW-SEPARABLE-EQUATION-007 Solve y'=y^2, y(0)=1.
- ZW-SEPARABLE-EQUATION-009 Solve y'=-x/y, y(0)=2.